This is just trailing zeroes

To find the trailing zeroes of each of the numbers, we use an iterative method of finding the total powers of five in each of the numbers.

We know that $2002$ < $5^{5}$ ,

so we get that the number of trailing zeroes of $2002!$ is

$\left \lfloor \frac {2002}{5} \right \rfloor + \left \lfloor \frac {2002}{5^{2}} \right \rfloor + \left \lfloor \frac {2002}{5^3} \right \rfloor + \left \lfloor \frac {2002}{5^4} \right \rfloor$

$\large = 400 + 80 + 16 + 3 = 499$

in a similar way, $1001 <\ 5^{5}$ the number of trailing zeroes of $1001!$ is

$\left \lfloor \frac {1001}{5} \right \rfloor + \left \lfloor \frac {1001}{5^{2}} \right \rfloor + \left \lfloor \frac {1001}{5^3} \right \rfloor + \left \lfloor \frac {1001}{5^4} \right \rfloor$

$\large = 200 + 40 + 8 + 1 = 249$

Thus, the total number of trailing zeroes of the expression is $499 - 2(249) = \large \boxed {1}$

Refer wiki and use the formula to calculate numerator &denominator. U will get 499/249 as the square of denominator isn't included. APPLY LOGIC:if we square 10 its result has twice trailing 0 then LHS i.e 10^2=100. Same is the case with 100^2&1000^2. HENCE IT'S GENERAL THAT R.H.S HAS TWICE THE TRAILING ZEROES THAN THE SQUARE OF L.H.S now let's calculate 1001!^2 It gives 249×2=498.&after putting its value in the original fraction i.e499/498 we get the nearest integer=1. CHEERS! ☆

that will be 499/498 = 1